You are correct that delta will not always be a fourth of epsilon.
–
The Chaz 2.0Apr 23 '12 at 2:24

4

The $\delta$ you pick depends on (i) the function $f$; (ii) the point $x_0$; and (iii) the $\epsilon$. Note that you want $f(x)$ to be within $\epsilon$ of $f(1)$, not of $1$. For this particular function, $\delta=\epsilon/4$ is not good enough.
–
Arturo MagidinApr 23 '12 at 2:25

3

If this were possible there would be no need for the whole $\delta$ vs. $\epsilon$ business$\ldots\ $.
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Christian BlatterApr 23 '12 at 14:30

It is a well-known canard that Freshmen believe all functions to be linear. For example, $1/(x+y) = 1/x + 1/y$ is obvious, no?
–
GEdgarJun 1 '12 at 13:55

1 Answer
1

If $f$ is linear, $f(x)=mx+b$ and you can take $\delta=\frac{\epsilon}{ |m|}$. You should plug this into the $\delta - \epsilon$ definition to see that it is true. For differentiable $f$, you can usually use something close to $\frac \epsilon {f'}$ (note that in the previous case $f'=m$) but you might need to use something smaller. This comes from the fact that the derivative gives the best local linear approximation, but higher order terms may be a problem.